HDU3117(快速幂)

Fibonacci Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1291    Accepted Submission(s): 545

Problem Description

The Fibonacci sequence is the sequence of numbers such that every element is equal to the sum of the two previous elements, except for the first two elements f0 and f1 which are respectively zero and one.

What is the numerical value of the nth Fibonacci number?

 

Input

For each test case, a line will contain an integer i between 0 and 10⁸ inclusively, for which you must compute the ith Fibonacci number fi. Fibonacci numbers get large pretty quickly, so whenever the answer has more than 8 digits, output only the first and last 4 digits of the answer, separating the two parts with an ellipsis (“...”).

There is no special way to denote the end of the of the input, simply stop when the standard input terminates (after the EOF).

 

Sample Input

0123453536373839406465

 

Sample Output

0112359227465149303522415781739088169632459861023...41551061...77231716...7565
 
 
//第四位运用矩阵的快速幂能够很好解决，但高四位就得结合数学知识，这个菜鸟不会
 
 
#include <cstdlib>#include <cstring>#include <cstdio>#include <iostream> #include<cmath>using namespace std;//N表示矩阵的N次幂int N;const int mod=10000;//此处假设矩阵为3*3阶//origin存放需计算的矩阵，res存放答案矩阵struct matrix{int a[2][2];}origin,res;//直接将2个矩阵相乘x*y,返回计算后的矩阵matrix multiply(matrix x,matrix y){matrix temp;memset(temp.a,0,sizeof(temp.a));for(int i=0;i<2;i++){for(int j=0;j<2;j++){for(int k=0;k<2;k++){temp.a[i][j]=(temp.a[i][j]+(x.a[i][k]*y.a[k][j])%mod)%mod;}}}return temp;}//将res初始化为单位矩阵，人为输入originvoid init(){origin.a[0][0]=origin.a[0][1]=origin.a[1][0]=1;origin.a[1][1]=0;//将res.a初始化为单位矩阵 memset(res.a,0,sizeof(res.a));res.a[0][0]=res.a[1][1]=1;                  }//矩阵快速幂的计算void calc(int n){while(n){if(n&1)res=multiply(res,origin);n>>=1;origin=multiply(origin,origin);}}void pre(int  n)  {         double a=sqrt(5.0);         double b=(sqrt(5.0)+1)/2;         double s=log(1.0/a)/log(10.0)+n*log(b)/log(10.0);         double t=s-(int)s+3;         double f=pow(10.0,t);         f=(int)f;         cout<<f<<"...";  }  int f[40];int main(){int i;f[0]=0;f[1]=1;for(i=2;i<40;i++)f[i]=f[i-1]+f[i-2];    while(cin>>N)    {if(N<40){printf("%d\n",f[N]);continue;}init();calc(N);pre(N);printf("%04d\n",res.a[1][0]);    }    return 0;}